## 67) is n divisible

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### 67) is n divisible

by ern5231 » Sun May 16, 2010 5:36 pm
X,Y,Z are consecutive integers. N = x + y + z . Is N divisible by 10?
(1) x + z is a multiple of 10
(2) y is a multiple of 10

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by achieve_dream » Sun May 16, 2010 6:42 pm

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by DeepthiRajan » Tue May 18, 2010 8:49 pm
IMO (B)

Reason: if x,y,z are 4,5,6 then 4+6 (x+z) is a multiple of 10 but 4+5+6=15 is not divisible by 10. Hence (1) is insufficient.

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by liferocks » Tue May 18, 2010 10:19 pm
X,Y,Z are consecutive integers i.e X,Y,Z are a-1,a,a+1..not in order

So N=3a..if we can definitely say a is a multiple of 10 or not question can be answered

From 1
X+Z=10k

this says X and Z has to be a+1 and a-1 and

a-1+a+1=10k or 2a=10k or a=5k...a may or may not be multiple of 10

option 2
Y is a multiple of 10..hence either of a-1,a,a+1 is a multiple of 10

if Y =a..then N is also multiple of 10 else if Y =a+1 or a-1..N is not a multiple of 10...insufficient

combining
Y =a and Y is a multiple of 10 hence N is multiple of 10

Ans option C
Please confirm the OA.
"If you don't know where you are going, any road will get you there."
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by gmatmachoman » Tue May 18, 2010 10:43 pm
ern5231 wrote:X,Y,Z are consecutive integers. N = x + y + z . Is N divisible by 10?
(1) x + z is a multiple of 10
(2) y is a multiple of 10
St 1 :
x + z is a multiple of 10

Let us plug in numbers:

x=19 y=20 z=21

x+z= 19+21 is a multiple of 10 .

And so N is a Divisible by 10.

St 2:y is a multiple of 10

Again plug in numbers: y=30
x=29,z=31

Now N= 29+30+31
N=90 which is divisible by 10.

IMO D

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by liferocks » Tue May 18, 2010 11:12 pm
gmatmachoman wrote:
ern5231 wrote:X,Y,Z are consecutive integers. N = x + y + z . Is N divisible by 10?
(1) x + z is a multiple of 10
(2) y is a multiple of 10
St 1 :
x + z is a multiple of 10

Let us plug in numbers:

x=19 y=20 z=21

x+z= 19+21 is a multiple of 10 .

And so N is a Divisible by 10.

St 2:y is a multiple of 10

Again plug in numbers: y=30
x=29,z=31

Now N= 29+30+31
N=90 which is divisible by 10.

IMO D
Hmm..I think you have missed few options

In condition 1

let us take x=4,y=5,z=6..x+z=10..multiple of 10
x+y+z=15..non multiple of 10

In condition 2
y=10
then x=9 z=11...or x=11,z=12 or x=8,z=9...

What say?
"If you don't know where you are going, any road will get you there."
Lewis Carroll

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by amitverma22 » Tue May 18, 2010 11:51 pm
Since X, Y, Z are consecutive integers

So Let Y = X + 1 and Z = X + 2

=> N = X + X + 1 + X + 2
N = 3(X + 1)

1) x + z is a multiple of 10 .
=> X + X + 2 is multiple of 10
=> 2(X+1) is multiple of 10
=> (X+1) is multiple of 5.

This means N is multiple of 5(and also 3) because N = 3(X+1)

2) y is a multiple of 10
=> (X+1) is multiple of 10

This certainly means N is multiple of 10 because N = 3(X+1)

So[spoiler] 2)[/spoiler] alone is sufficient hence IMO ans is B

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by nikhilkatira » Wed May 19, 2010 8:28 pm
+1 for B
Best,
Nikhil H. Katira

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by krazy800 » Wed May 19, 2010 8:36 pm
IMO B
Aiming High

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by [email protected] » Thu May 20, 2010 1:37 am
amitverma22 wrote:Since X, Y, Z are consecutive integers

So Let Y = X + 1 and Z = X + 2

=> N = X + X + 1 + X + 2
N = 3(X + 1)

1) x + z is a multiple of 10 .
=> X + X + 2 is multiple of 10
=> 2(X+1) is multiple of 10
=> (X+1) is multiple of 5.

This means N is multiple of 5(and also 3) because N = 3(X+1)

2) y is a multiple of 10
=> (X+1) is multiple of 10

This certainly means N is multiple of 10 because N = 3(X+1)

So[spoiler] 2)[/spoiler] alone is sufficient hence IMO ans is B

let x = m, then y = m+1 and z = m+2

implying that N = x+y+z = 3m + 3 or 3(m+1)

is 3m + 3 = 10k?

1) x+z = 10r implying 2m+2 = 10r insufficient!
2) y = 10q implying m+1 = 10q insufficient!

combine 1 and 2: x+z + y = (2m+2) + (m+1) = 10r+10q
this in essence means 3m+3 = 10 (r+q)

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by amitverma22 » Thu May 20, 2010 4:05 am
[email protected] wrote:
amitverma22 wrote:Since X, Y, Z are consecutive integers

So Let Y = X + 1 and Z = X + 2

=> N = X + X + 1 + X + 2
N = 3(X + 1)

1) x + z is a multiple of 10 .
=> X + X + 2 is multiple of 10
=> 2(X+1) is multiple of 10
=> (X+1) is multiple of 5.

This means N is multiple of 5(and also 3) because N = 3(X+1)

2) y is a multiple of 10
=> (X+1) is multiple of 10

This certainly means N is multiple of 10 because N = 3(X+1)

So[spoiler] 2)[/spoiler] alone is sufficient hence IMO ans is B

let x = m, then y = m+1 and z = m+2

implying that N = x+y+z = 3m + 3 or 3(m+1)

is 3m + 3 = 10k?

1) x+z = 10r implying 2m+2 = 10r insufficient!
2) y = 10q implying m+1 = 10q insufficient!

combine 1 and 2: x+z + y = (2m+2) + (m+1) = 10r+10q
this in essence means 3m+3 = 10 (r+q)

Taking your case 2)

y = 10q implying m+1 = 10q
This implies 3(m + 1) = 3*10q
This implies 3m + 3 = 10*(3q)

Hence 3m + 3 is definitely multiple of 10.
So 2) is sufficient to answer the question

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by [email protected] » Thu May 20, 2010 5:06 am
amitverma22 wrote:
[email protected] wrote:
amitverma22 wrote:Since X, Y, Z are consecutive integers

So Let Y = X + 1 and Z = X + 2

=> N = X + X + 1 + X + 2
N = 3(X + 1)

1) x + z is a multiple of 10 .
=> X + X + 2 is multiple of 10
=> 2(X+1) is multiple of 10
=> (X+1) is multiple of 5.

This means N is multiple of 5(and also 3) because N = 3(X+1)

2) y is a multiple of 10
=> (X+1) is multiple of 10

This certainly means N is multiple of 10 because N = 3(X+1)

So[spoiler] 2)[/spoiler] alone is sufficient hence IMO ans is B

let x = m, then y = m+1 and z = m+2

implying that N = x+y+z = 3m + 3 or 3(m+1)

is 3m + 3 = 10k?

1) x+z = 10r implying 2m+2 = 10r insufficient!
2) y = 10q implying m+1 = 10q insufficient!

combine 1 and 2: x+z + y = (2m+2) + (m+1) = 10r+10q
this in essence means 3m+3 = 10 (r+q)

Taking your case 2)

y = 10q implying m+1 = 10q
This implies 3(m + 1) = 3*10q
This implies 3m + 3 = 10*(3q)

Hence 3m + 3 is definitely multiple of 10.
So 2) is sufficient to answer the question
i see; i agree with you.

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by kstv » Thu May 20, 2010 7:43 am
ern5231 wrote:X,Y,Z are consecutive integers. N = x + y + z . Is N divisible by 10?
(1) x + z is a multiple of 10
(2) y is a multiple of 10
x,y & z are consec nos.
so 2y = x+z

1) x+z = 10a , y = 5a
N= 15a divisible by 5 not necessarily 10 . Insuff

2) y = 10 a , x+z = 20a
N = 30a so it is divisible by 10. Suff

IMO B

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by layzzer » Thu May 20, 2010 12:51 pm
I'm reposting the following

In condition 1

let us take x=4,y=5,z=6..x+z=10..multiple of 10
x+y+z=15..non multiple of 10

In condition 2
y=10
then x=9 z=11...or x=11,z=12 or x=8,z=9...

What say?

My question then,
now when a questions states numbers x, y, z are consecutive numbers, does it necessarily mean that x is the smallest, y is the middle number and z is the largest?

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by kstv » Thu May 20, 2010 8:48 pm
layzzer wrote:I'm reposting the following
My question then,
now when a questions states numbers x, y, z are consecutive numbers, does it necessarily mean that x is the smallest, y is the middle number and z is the largest?
Good point
1) x+z is a multiple of 10
since sum of two consecutive nos cannot be multiple of 10, x and z are not consecutive nos. So x and z are the smallest and the largest nos. in the series or vice versa. But it is not Suff to say whether N is a multiple of 10.
2) y is multiple of 10. but y can be the first , middle or the last number. Only if y is the middle value than N is a multiple of 10. Insufficient.
Combining it is possible to deduce that y is the middle no. Rest is the same.
OK IMHO it is C.